Composition relation of R1 ∘ R2

$$1\to1\to1$$ $$2\to2\to3$$ $$2\to2\to4$$ $$2\to3\to5$$ $$2\to5\to2$$ $$4\to3\to5$$ $$5\to5\to5$$

But, as pointed out by @Rag in the comments, there is one additional:

$$5\to5\to2$$

And, as pointed out by @MithleshUpadhyay, a second way to obtain $(2,5)$ is $$ 2\to5\to5$$

An alternative is through matrix representations of relations ($a_{ij}=1$ if $(i,j)$ is present in the relation, $0$ otherwise) with composition of relations replaced by matrix product (in the same order as in the composition, with boolean addition convention: $1+1=1$). This can be very useful on a computer. Here, we have :

$$\begin{pmatrix}1&0&0&0&0\\0&1&1&0&1\\0&0&0&0&0\\0&0&1&0&0\\0&0&0&0&1\end{pmatrix} \times \begin{pmatrix}1&0&0&0&0\\0&0&1&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&1&0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0&0&0\\0&1&1&1&1\\0&0&0&0&0\\0&0&0&0&1\\0&1&0&0&1\end{pmatrix}$$